\(\int (d x)^m (a+b \text {arctanh}(\frac {c}{x^2})) \, dx\) [183]

Optimal result

Integrand size = 16, antiderivative size = 75 \[ \int (d x)^m \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {(d x)^{1+m} \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )}{d (1+m)}-\frac {2 b c d (d x)^{-1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1-m}{4},\frac {5-m}{4},\frac {c^2}{x^4}\right )}{1-m^2} \] Output:

(d*x)^(1+m)*(a+b*arctanh(c/x^2))/d/(1+m)-2*b*c*d*(d*x)^(-1+m)*hypergeom([1 
, 1/4-1/4*m],[5/4-1/4*m],c^2/x^4)/(-m^2+1)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.91 \[ \int (d x)^m \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {(d x)^m \left ((-1+m) x^2 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )+2 b c \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4}-\frac {m}{4},\frac {5}{4}-\frac {m}{4},\frac {c^2}{x^4}\right )\right )}{(-1+m) (1+m) x} \] Input:

Integrate[(d*x)^m*(a + b*ArcTanh[c/x^2]),x]
 

Output:

((d*x)^m*((-1 + m)*x^2*(a + b*ArcTanh[c/x^2]) + 2*b*c*Hypergeometric2F1[1, 
 1/4 - m/4, 5/4 - m/4, c^2/x^4]))/((-1 + m)*(1 + m)*x)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6464, 862, 888}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (d x)^m \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx\)

\(\Big \downarrow \) 6464

\(\displaystyle \frac {2 b c d^2 \int \frac {(d x)^{m-2}}{1-\frac {c^2}{x^4}}dx}{m+1}+\frac {(d x)^{m+1} \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )}{d (m+1)}\)

\(\Big \downarrow \) 862

\(\displaystyle \frac {(d x)^{m+1} \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )}{d (m+1)}-\frac {2 b c d \left (\frac {1}{x}\right )^{m-1} (d x)^{m-1} \int \frac {\left (\frac {1}{x}\right )^{-m}}{1-\frac {c^2}{x^4}}d\frac {1}{x}}{m+1}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {(d x)^{m+1} \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )}{d (m+1)}-\frac {2 b c d (d x)^{m-1} \operatorname {Hypergeometric2F1}\left (1,\frac {1-m}{4},\frac {5-m}{4},\frac {c^2}{x^4}\right )}{(1-m) (m+1)}\)

Input:

Int[(d*x)^m*(a + b*ArcTanh[c/x^2]),x]
 

Output:

((d*x)^(1 + m)*(a + b*ArcTanh[c/x^2]))/(d*(1 + m)) - (2*b*c*d*(d*x)^(-1 + 
m)*Hypergeometric2F1[1, (1 - m)/4, (5 - m)/4, c^2/x^4])/((1 - m)*(1 + m))
 

Defintions of rubi rules used

rule 862

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-c^ 
(-1))*(c*x)^(m + 1)*(1/x)^(m + 1)   Subst[Int[(a + b/x^n)^p/x^(m + 2), x], 
x, 1/x], x] /; FreeQ[{a, b, c, m, p}, x] && ILtQ[n, 0] &&  !RationalQ[m]
 

rule 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 6464

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))*((d_)*(x_))^(m_), x_Symbol] : 
> Simp[(d*x)^(m + 1)*((a + b*ArcTanh[c*x^n])/(d*(m + 1))), x] - Simp[b*c*(n 
/(d^n*(m + 1)))   Int[(d*x)^(m + n)/(1 - c^2*x^(2*n)), x], x] /; FreeQ[{a, 
b, c, d, m, n}, x] && IntegerQ[n] && NeQ[m, -1]
 

Maple [F]

\[\int \left (d x \right )^{m} \left (a +b \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )\right )d x\]

Input:

int((d*x)^m*(a+b*arctanh(c/x^2)),x)
 

Output:

int((d*x)^m*(a+b*arctanh(c/x^2)),x)
 

Fricas [F]

\[ \int (d x)^m \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\int { {\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )} \left (d x\right )^{m} \,d x } \] Input:

integrate((d*x)^m*(a+b*arctanh(c/x^2)),x, algorithm="fricas")
 

Output:

integral((b*arctanh(c/x^2) + a)*(d*x)^m, x)
 

Sympy [F]

\[ \int (d x)^m \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\int \left (d x\right )^{m} \left (a + b \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}\right )\, dx \] Input:

integrate((d*x)**m*(a+b*atanh(c/x**2)),x)
 

Output:

Integral((d*x)**m*(a + b*atanh(c/x**2)), x)
 

Maxima [F]

\[ \int (d x)^m \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\int { {\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )} \left (d x\right )^{m} \,d x } \] Input:

integrate((d*x)^m*(a+b*arctanh(c/x^2)),x, algorithm="maxima")
 

Output:

1/2*(4*c*d^m*integrate(x^2*x^m/((m + 1)*x^4 - c^2*(m + 1)), x) + (d^m*x*x^ 
m*log(x^2 + c) - d^m*x*x^m*log(x^2 - c))/(m + 1))*b + (d*x)^(m + 1)*a/(d*( 
m + 1))
 

Giac [F]

\[ \int (d x)^m \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\int { {\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )} \left (d x\right )^{m} \,d x } \] Input:

integrate((d*x)^m*(a+b*arctanh(c/x^2)),x, algorithm="giac")
 

Output:

integrate((b*arctanh(c/x^2) + a)*(d*x)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d x)^m \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\int {\left (d\,x\right )}^m\,\left (a+b\,\mathrm {atanh}\left (\frac {c}{x^2}\right )\right ) \,d x \] Input:

int((d*x)^m*(a + b*atanh(c/x^2)),x)
 

Output:

int((d*x)^m*(a + b*atanh(c/x^2)), x)
 

Reduce [F]

\[ \int (d x)^m \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {d^{m} \left (x^{m} a x +\left (\int x^{m} \mathit {atanh} \left (\frac {c}{x^{2}}\right )d x \right ) b m +\left (\int x^{m} \mathit {atanh} \left (\frac {c}{x^{2}}\right )d x \right ) b \right )}{m +1} \] Input:

int((d*x)^m*(a+b*atanh(c/x^2)),x)
 

Output:

(d**m*(x**m*a*x + int(x**m*atanh(c/x**2),x)*b*m + int(x**m*atanh(c/x**2),x 
)*b))/(m + 1)